Question

A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20 degrees above the ground from a height of 4.5 ft. Approximately how long does it take for the ball to hit the ground? Acceleration due to gravity is 32 ft./s^2.

A. 2.35 seconds
B. 2.47 seconds
C. 6.46 seconds
D. 6.50 seconds

Answer 1

T= 2.35 seconds

Explanation:

⇒The question is on the time of flight.

⇒Time of flight is the time taken for a projected object to reach the ground.It depends on the projectile angle and the initial velocity of the projectile

Given;

Initial velocity of ball= 110ft./sec.

The projectile angle= 20°

Acceleration due to gravity, g=32 ft./s²

⇒Formulae for time of fright T= (2×u×sin Ф)/g

Where T=time of fright, u=initial velocity of projectile, Ф=projectile angle and g=acceleration due o gravity.

Substituting values

T= (2×u×sin Ф)/g

T=( 2×110×sin 20°) / 32

T= 2.35 seconds

Answer 2

Option B - Time taken for the ball to hit the ground is 2.47 seconds.

Explanation:

Given : A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20 degrees above the ground from a height of 4.5 ft.

To find : How long does it take for the ball to hit the ground?

Solution :

According to question,

The equation that models the height of the ball in feet as a function of time is

`h(t) = h_0 + v_0t -16t^2`

Where,
`h_0` is the initial height,

`v_0` is the initial velocity and

t is the time in seconds.

We have given,

Initial height,
`h_0=4.5 ft.`

A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20 degrees.

The initial speed,
`v_0=110* \sin(206\circ)`

`v_0=37.62ft/s`

We have to find the time for the ball to hit the ground i.e. h(t)=0

Substitute all the values in the formula,

`0 =4.5+ 37.62t -16t^2`

Applying quadratic formula to solve the equation,

The solution of quadratic general equation
`ax^2+bc+c=0` is

`x=(-b\pm√(b^2-4ac))/(2a)`

Where, a=-16 , b=37.62 , c=4.5

Substituting in the formula,

`t=(-37.62\±√((37.62)^2 -4(-16)(4.5)))/(2(-16))`

`t=(-37.62\±√(1703.2644))/(-32)`

`t=(-37.62\±41.270)/(-32)`

`t=(-37.62+41.270)/(-32),(-37.62-41.270)/(-32)`

`t=-0.114,2.47`

neglecting the negative value

t=2.47 seconds

Therefore,Option B is correct.

Time taken for the ball to hit the ground is 2.47 seconds.